# Question #fef55

Feb 5, 2015

The reaction's percent yield is $\text{90.5%}$.

So, start with the balanced chemical equation for the decomposition of $C a C {O}_{3}$

$C a C {O}_{3 \left(s\right)} \to C a {O}_{\left(s\right)} + C {O}_{2 \left(g\right)}$

Notice that you've got $\text{1:1}$ mole ratios between all the compounds involved in the reaction. This is important because you'll use the mole ratio $C a C {O}_{3}$ has with $C a O$ to determine what the theoretical yield of the reaction is.

The number of $C a C {O}_{3}$ moles you have is

$2.00 \cdot {10}^{3} \text{g" * ("1 mole")/("100.1 g") = "20.0 moles}$

This means that the mass of $C a O$ produced if the reaction had a 100% yield would be

$\text{20.0 moles" * ("58.1 g")/("1 mole") = "1162 g" = 1.16 * 10^(3)"g}$

Since percent yield is defined as actual yield divided by theoretical yield and multiplied by 100, you'd get

$\text{% yield" = "actual"/"theoretic" * 100 = (1.05 * 10^(3)"g")/(1.16 * 10^(3)"g") * 100 = "90.5%}$

You can double check the result by using moles instead of grams. You know that you have $\text{20 moles}$ of $C a O$ produced in theory. The actual number of moles of $C a O$ produced is

$1.05 \cdot {10}^{3} \text{g" * ("1 mole")/("58.1 g") = "18.1 moles}$

Therefore, the percent yield is one again

$\text{% yield" = ("18.1 moles CaO")/("20.0 moles CaO") * 100 = "90.5%}$